Evaluate:
\sum_{r=0}^{n}{^{n}C_{r}\left(r-nx \right)^{2}x^{r}\left(1-x \right)^{n-r}}
21after splitting the square term,we have to find
1) S = 2nx\sumnCr r xr(1-x)n-r
which may be done by considering the expression P=(1-x + y )n
if we put x= y in the expression T= ydPdy ,we get S2nx
2)S1 = \sumnCr r2 xr (1-x)n-r
S1 is given by putting x= y in the expression ydTdy
3)S3 = n2x2\sumnCr xr (1-x)n-r
we can have S3 by putting x=y in P, so S3 is simply 1
the required term is given by S1 - S + S3
1708\bf(1+y)^n=\sum_{r=0}^{n}^nC_{r}.y^r$\\\\ Now $\bf n.(1+y)^{n-1}=\sum_{r=0}^{n}r.^nC_{r}.y^{r-1}$\\\\ Again $\bf n.(n-1)(1+y)^{n-2}=\sum_{r=0}^{n}r.(r-1)^nC_{r}.y^{r-2}$\\\\ Now $\bf \sum_{r=0}^{n}(r-nx)^2.^nC_{r}.x^r.(1-x)^{n-r}$\\\\ $\bf= \sum_{r=0}^{n}(\underbrace{\bf r^2-r}+\underbrace{\bf r-2r.n.x}+n^2x^2).^nC_{r}.x^r.(1-x)^{n-r}$\\\\ $\bf=(1-x)^n\left\{\sum_{r=0}^{n}^nC_{r}.r(r-1).y^r+(1-2nx).\sum_{r=0}^{n}^nC_{r}.r.y^r+n^2.x^2.\sum_{r=0}^{n}^nC_{r}.y^r \right\}$\\\\ Where $\bf y=\frac{x}{1-x}$\\\\ $\bf=(1-x)^n.\left\{n.(n-1).(1+y)^{n-2}.y^2+(1-2nx).n.(1+y)^{n-1}.y+n^2.x^2.(1+y)^n \right\}$\\\\ After Simplifying, We get.....\\\\ $\bf =nx.(1-x)$\\\\
